Re: [PATCH v4 3/3] sched: optimize migration by forcing rmb() and updating to be called once

Peter Zijlstra
2015-11-17T23:55:22Z

On Wed, Nov 18, 2015 at 08:37:00AM +0900, Byungchul Park wrote:

> Which one do you think to be fixed? The one above migrate_task_rq_fair()?
> I wonder if it would be ok even it does not hold pi_lock in
> migrate_task_rq_fair(). If you say *no problem*, I will try to fix the
> comment.

The one above migrate_task_rq_fair() is obviously broken, as
demonstrated by the move_queued_task() case.

Also, pretty much all runnable task migration code will not take
pi_lock, see also {pull,push}_{rt,dl}_task().

Note that this is done very much by design, task_rq_lock() is the thing
that fully serializes a task's scheduler state. Runnable tasks use
rq->lock, waking tasks use pi_lock.

> > I meant, if you call __set_task_cpu() before
> > sched_class::migrate_task_rq(), in that case task_rq_lock() will no
> > longer fully serialize against set_task_cpu().
> >
> > Because once you've called __set_task_cpu(), task_rq_lock() will acquire
> > the _other_ rq->lock. And we cannot rely on our rq->lock to serialize
> > things.
>
> I agree with you if migtrate_task_rq() can be serialized by rq->lock
> without holding pi_lock. (even though I am still wondering..)

move_queued_task() illustrates this.

> But I thought it was no problem if migrate_task_rq() was serialized only
> by pi_lock as the comment above the migrate_task_rq() describes, because
> breaking rq->lock does not affect the sericalization by pi_lock.

Right, but per the above, we cannot assume pi_lock is in fact held over
this.